Question

我正在进行A-B测试，看看哪种样式的网页可能会获得更多订单的产品，并试图将一个页面称为“更好”，因为它更频繁地显示。

我的MYSQL数据库中有8列。

Orders_Page_One
Orders_Page_Two
Orders_Page_Three
Visitors_Page_One
Visitors_Page_Two
Visitors_Page_Three
Total_Orders
Total_Visitors

目前我只运行查询：


$result = mysql_query("SELECT *, (Total_Orders / Total_Visitors) AS order_percent FROM orders WHERE category = clothes ORDER BY order_percent DESC LIMIT 5") or die(mysql_error());

这可行但

这允许我显示哪些订单最受欢迎，但不考虑它们出现在哪个页面上。

我试过了：


$result = mysql_query("SELECT *, (Orders_Page_One / Visitors_Page_One) + (Orders_Page_Two / Visitors_Page_Two) + (Orders_Page_Three / Visitors_Page_Three) AS order_percent FROM orders WHERE category = clothes ORDER BY order_percent DESC LIMIT 5") or die(mysql_error());

但这似乎没有给出准确的结果。

我也试过


$result = mysql_query("SELECT *, ((Orders_Page_One / Visitors_Page_One) + (Orders_Page_Two / Visitors_Page_Two) + (Orders_Page_Three / Visitors_Page_Three)) AS order_percent FROM orders WHERE category = clothes ORDER BY order_percent DESC LIMIT 5") or die(mysql_error());

但这会产生语法错误。

假设：

Orders Page One = 10 and Visitors Page One = 20 
Orders Page Two = 10 and Visitors Page Two = 40 
Orders Page Three = 10 and Visitors Page Three = 50

我想要的数学结果是：

(10 / 20) + (10 / 40) + (10 / 50)

OR

0.5 + 0.25 + 0.20 = 0.95

有什么建议吗？我想我只是没有正确设置数学。

经过多次阅读后，我想我需要使用多项选择：

IE：


    Select (orders_page_1/visitors_page_1) as order_percent_1 from orders
    Select (orders_page_2/visitors_page_2) as order_percent_2 from orders
    Select (orders_page_3/visitors_page_3) as order_percent_3 from orders
    Select (order_percent_1 + order_percent_two + order_percent_3) as order_percent
    WHERE category = clothes
    ORDER BY order_percent DESC

但格式不确定？或者，如果这可行？

$result = mysql_query("

SELECT * FROM orders, (
    SELECT (orders_page_1 / visitors_page_one) AS ord1 FROM orders WHERE category = 'clothing' ORDER BY ord1 DESC LIMIT 2) AS ord_1,
      (
    SELECT (orders_page_2 / visitors_page_two) AS ord2 FROM orders WHERE category = 'clothing' ORDER BY ord2 DESC LIMIT 2) AS ord_2,
      (
    SELECT (orders_page_3 / visitors_page_3) AS ord3 FROM orders WHERE category = 'clothing' ORDER BY ord3 DESC LIMIT 2) AS ord_3,

    ORDER BY (ord_1 +ord_2 + ord_3) DESC LIMIT 2") or die(mysql_error());

然而，这是错误：

您的SQL语法有错误;检查与MySQL服务器版本对应的手册，以便在第8行的“ORDER BY（ord_1 + ord_2 + ord_3）DESC LIMIT 2”附近使用正确的语法

Answer 1

一个可能的问题是

WHERE category = clothes

应该是

WHERE category = 'clothes'

立即尝试，看看你得到了什么。

MYSQL中的乘法，除法和加法

1 个答案: